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STAT 408 Spring 2009
Homework
(due Friday, February 20, by 3:00 p.m.)
Be sure to show all your work; your partial credit might depend on it.
No credit will be given without supporting work.
1. Sally sells seashells by the seashore. The daily sales X of the seashells have the
following probability distribution:
x f ( x ) x ⋅ f ( x ) ( x − μ) 2 ⋅ f ( x ) 0 1 2 3 4
a) Find the missing probability f ( 0 ) = P ( X = 0 ).
f ( 0 ) = 1 − [0.25 + 0.20 + 0.15 + 0.10] = 1 − 0.70 = 0..
b) Find the probability that she sells at least three seashells on a given day.
P(X ≥ 3) = P(X = 3) + P(X = 4) = 0.15 + 0.10 = 0..
c) Find the expected daily sales of seashells.
E(X) = μ (^) X = (^) ∑ ⋅ x
x f x
all
d) Find the standard deviation of daily sales of seashells.
SD(X) = σX = (^ )^ ( ) all
2 ∑ −μX ⋅ x
x f x = 1. 75 ≈ 1..
1. (continued)
Suppose each shell sells for $5.00. However, Sally must pay $3.00 daily for the
permit to sell shells. Let Y denote Sally’s daily profit. Then Y = 5 ⋅ X – 3.
e) Find the probability that Sally’s daily profit will be at least $10.00. [ Hint: How
many shells must Sally sell to have daily profit of at least $10.00? ]
x y f ( x ) = f ( y )
P(Y ≥ 10) = P(X ≥ 3) = 0..
( Sally’s daily profit is at least $10.00 if and only if she sells at least 3 shells. )
f) Find Sally’s expected daily profit.
μ Y = E(Y) = $5.00 ⋅ E(X) − $3.00 = $5.00 ⋅ 1.50 − $3.00 = $ 4..
( On average, Sally sells 1.5 shells per day, her expected revenue is $7.50. Her expected
profit is $4.50 since she has to pay $3.00 for the permit. )
OR
x y f ( x ) = f ( y ) y^ ⋅^ f^ ( y )
μ (^) Y = E(Y) = (^) ∑ ⋅ y
y f y
all
3. What is the probability that a 5-card poker hand drawn from a standard 52-card deck
has …
a) “four of a kind” [ 4 cards of the same denomination (value), plus 1 additional card ]?
[ Hint: What is the probability that a 5-card poker hand drawn from a standard
52-card deck has four Aces? four Kings? … four 2’s? ]
A’s other
C
C C
⇒ P ( 4 A’s ) = P ( 4 K’s ) = P ( 4 Q’s ) = … = P ( 4 3 ’s ) = P ( 4 2 ’s ) =
⇒ P ( “Four of a kind” ) = 13 ×
OR
choose the value for the “four of a kind”
take 4 out of 4 cards of chosen value
choose one more card from the remaining 48
13 C^1 ×^4 C^4 ×^48 C^1 (^13) × 1 × 48 = 624
C
b) “full house” [ 3 cards of one denomination, plus 2 cards of a second denomination ]?
choose the value for the “three of a kind”
take 3 out of 4 cards of chosen value
choose the value for the “two of a kind”
take 2 out of 4 cards of chosen value
13 C^1 ×^4 C^3 ×^12 C^1 ×^4 C^2
13 × 4 × 12 × 6 = 3,
C
c) “straight” [ 5 cards of consecutive denominations (including possibly all 5 cards of the
same suit – straight flush, and 10 J Q K A of the same suit – royal flush) ]?
[ Hint: A 2 3 4 5 through 10 J Q K A. ]
A 2 3 4 5 through 10 J Q K A – 10 possible choices for the values of the 5 cards.
choose the values for the 5 cards
choose the suit for each of the 5 cards
10 C^1 ×^ (^4 C^1 )^
5
(^10) × 1,024 = 10,
C
Note: A 2 3 4 5 through 9 10 J Q K of the same suit ( 36 possibilities ) – straight flush
10 J Q K A of the same suit ( 4 possibilities ) – royal flush
⇒ 10,200 possible ways to have a “straight” ( just a “straight” )
36 possible ways to have a “straight flush”
4 possible ways to have a “royal flush”
d) “flush” [ any 5 cards all of the same suit (including royal flush and straight flush) ]?
choose the suit choose 5 cards out 13 in the chosen suit
4 C^1 ×^13 C^5 (^4) × 1,287 = 5,
C
Note: 5,108 possible ways to have a “flush” ( just a “flush” )
36 possible ways to have a “straight flush”
4 possible ways to have a “royal flush”
5. a) Alex takes a multiple choice quiz in his Anthropology 100 class. The quiz has
10 questions, each has 4 possible answers, only one of which is correct. Alex did not study for the quiz, so he guesses independently on every question. What is the probability that Alex answers exactly 2 questions correctly?
X = number of questions Alex answers correctly. Binomial, n = 10, p =^1 / 4 = 0.25.
P ( X = 2 ) = 10 C (^) 2 ⋅( 0. 25 ) 2 ⋅( 0. 75 )^8 ≈ 0..
b) Alex takes a quiz in his Anthropology 100 class. The quiz consists of 10, questions the first 4 are True-False, the last 6 are multiple choice questions with 4 possible answers each, only one of which is correct. Alex did not study for the quiz, so he guesses independently on each question. Find the probability that he answers exactly 2 questions correctly.
Let X = the number of True-False questions answered correctly, Y = the number of multiple choice questions answered correctly.
X has Binomial distribution, n 1 = 4, p 1 = 0.50.
Y has Binomial distribution, n 2 = 6, p 2 = 0.25.
P( X + Y = 2 ) = P( X = 0 ∩ Y = 2 ) + P( X = 1 ∩ Y = 1 ) + P( X = 2 ∩ Y = 0 )
= P( X = 0 ) × P( Y = 2 ) + P( X = 1 ) × P( Y = 1 ) + P( X = 2 ) × P( Y = 0 )
= 4 C 0 ⋅ 0. 500 ⋅ 0. 504 × 6 C 2 ⋅ 0. 252 ⋅ 0. 754
+ 4 C 1 ⋅ 0. 501 ⋅ 0. 503 × 6 C 1 ⋅ 0. 251 ⋅ 0. 755
+ 4 C 2 ⋅ 0. 502 ⋅ 0. 502 × 6 C 0 ⋅ 0. 250 ⋅ 0. 756
From the textbook:
2.2-2 Must have 1 = 0.9 +
c + c + c + c + c + c.
⇒ c =
E ( Payment ) =
1 ⋅ c^ + ⋅ c + ⋅ c + ⋅ c + ⋅ c =
2.2-4 E ( X ) = ( ) ( ) ( )
E ( X 2 ) = ( ) ( ) ( )
− 1 2 ⋅ 4 +^2 ⋅ + −^2 ⋅ = ;
E ( 3 X 2 – 2 X + 4 ) = ( )
3 0 4^20
⋅ ^ ⋅.
2.3-4 E X^1 E( X ) 1 ( E( X) ) 1 ( μ μ)
σ
μ σ
μ σ σ
μ (^) − − −
−^ = = = = 0 ;
2 [^ (^ )^2 ]^22
E X^1 E X σ σ
μ σ σ
−^ = 1.
2.4-4 (a) P(X ≤ 5) = 0.5269 ;
(b) P(X ≥ 6) = 1 – P(X ≤ 5) = 0.4731 ;
(c) P(X = 7) = P(X ≤ 7) – P(X ≤ 6) = 0.8883 – 0.7393 = 0.1490 ;
(d) μ = 12 (0.45) = 5.4 , σ 2 = 12 (0.45)(0.55) = 2.97 , σ = 2. 97 = 1..
2.4-8 (a) X is b (15, 0.2), Binomial distribution with n = 15 and p = 0.2;
(b) μ = 15 (0.2) = 3 , σ 2 = 15 (0.2)(0.8) = 2.4 , σ = 2. 4 = 1.549 ;
(c) P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.8358 = 0..
